ch02-p062 - 62. The graph shows y = 25 m to be the highest...

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(a) To find the acceleration due to gravity g p on that planet, we use Eq. 2-15 (with + y up) () ( ) 2 2 0 11 25 m 0 0 2.5 s 2.5 s 22 pp yy v t g t g −= + ¡ + so that g p = 8.0 m/s 2 . (b) That same (max) point on the graph can be used to find the initial velocity. ( ) 00 0 25 m 0 0 2.5 s yy v v t v + ¡ + Therefore, v 0 = 20 m/s. 62. The graph shows
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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